# SOLUTION: i posted these problems 4 days ago,but haven't received a response.could someone please help me.Thanks. sketch the graphs,using the roots to find the function and the turning po

Algebra ->  Algebra  -> Graphs -> SOLUTION: i posted these problems 4 days ago,but haven't received a response.could someone please help me.Thanks. sketch the graphs,using the roots to find the function and the turning po      Log On

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 Question 207794: i posted these problems 4 days ago,but haven't received a response.could someone please help me.Thanks. sketch the graphs,using the roots to find the function and the turning point 1.x = -8, x = 4 2.x = -3, x = 3 thanks for the help.hagd Answer by vleith(2825)   (Show Source): You can put this solution on YOUR website!You are given roots, so you can figure out the quadratic. I will do the first one, you can use the same process to do the second one 1) roots are x=-8 and x=4 You know that at the roots, the value of the function is 0 You also know that for any product to equal zero, one or more terms in that function must be 0. Thus x =-8 x+8 = 0 x= 4 x-4=0 Now you have the two terms for your equation This is the equation for the first problem. Since is it q quadratic, you know it is a parabola that opens up (opens up since the x^2 term is positive). You also know that the 'turning point' will be halfway between the two root. Since the roots are 12 units apart, the 'turning point' will be 6 units from each root. -8+6 = -2 4-6 = -2 So find the value of f(x) when x=-2 and you'll have the minimum. So the 'turning point is the point (2,-20) Plot the funtion here --> http://www.wolframalpha.com/input/?i=x^2+%2B+4x+-+32+ Now do the same process for the other one. I am sure you can do it. If you still have questions, drop me an email